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PSLE

PSLE Foundation Mathematics

<p>In the diagram, \(AB\) is a diameter of the circle with centre \(O\). \(DE\) and \(BF\) are tangents to the circle at \(C\) and \(B\) respectively. \(DCE\) and \(BEF\) are straight lines. Prove that</p>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1782643825419-7qc4gmhh4c5.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3ODI2NDM4MjU0MTktN3FjNGdtaGg0YzUucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTYzNiwiZXhwIjoxNzkyNzI1OTM2fQ.qHVIi5SO3zjxXEWxTYEQCVk8Xa5zWtUbRFc-XpRBHJc" style="max-width: min(180px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-content"><span class="list-item-text">\(\Delta ABC\) and \(\Delta BFC\) are similar,</span><span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-content"><span class="list-item-text">hence or otherwise, show that&nbsp;\(\Delta CFE\) is an isosceles triangle.</span><span class="inline-marks">[4]</span></div></li></ol>

In the diagram, \(AB\) is a diameter of the circle with centre \(O\). \(DE\) and \(BF\) are tangents to the circle at \(C\) and \(B\) respectively. \(DCE\) and \(BEF\) are straight lines. Prove that

\(\triangle ABC\) and \(\triangle BFC\) are similar,

hence or otherwise, show that \(\triangle CEF\) is an isosceles triangle.

Alternate Segment Theorem

Tangents Perpendicular to Radii

Similarity

Angles in a Semicircle and the Converse

Equal Tangents from an External Point

2025 AHS PRELIMS P1 Q13

Proofs in Plane Geometry

<p>In the diagram, \(TA\) and \(TB\) are tangents from \(T\) to the circle with centre \(O\) at \(A\) and \(B\), respectively. \(AC\) is a diameter of the circle. Prove that</p>
<div class="graph-block" data-graph-config="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" data-width="198" data-height="163" data-proportional style="width:198px;height:163px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:198px;--graph-aspect:198 / 163;"></div>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">angle \(AOB = 2 \times\) angle \(ACB\),</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">\(CB\) is parallel to \(OT\),</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">triangles \(OBC\) and \(TAB\) are similar, and</div></li><li data-list-item-id="item-iv" data-list-item-path="item-iv"><div class="list-item-paragraph">quadrilateral \(AOBT\) is a cyclic quadrilateral, i.e., there exists a circle that passes through \(A\), \(O\), \(B\) and \(T\).</div></li></ol>

In the diagram, \(TA\) and \(TB\) are tangents from \(T\) to the circle with centre \(O\) at \(A\) and \(B\), respectively. \(AC\) is a diameter of the circle. Prove that

triangles \(OBC\) and \(TAB\) are similar, and

quadrilateral \(AOBT\) is a cyclic quadrilateral, i.e., there exists a circle that passes through \(A\), \(O\), \(B\) and \(T\).

Angles at the Centre and Circumference

Proving Concyclicity

Angles Formed by Two Parallel Lines and a Transversal

<p>In the diagram, \(TAR\) is the tangent to the circle at \(A\). \(DCA\) and \(DOB\) are straight lines. It is given that angle \(BAR = 3 \times \text{angle}\hspace{0.5em} OCB\), prove that</p>
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<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">angle \(ODC = 2 \times \text{angle}\hspace{0.5em} OBC\),</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">triangle \(COD\) is an isosceles triangle, and</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">angle \(DCO = \text{angle}\hspace{0.5em} CAT + 90^\circ\).</div></li></ol>

In the diagram, \(TAR\) is the tangent to the circle at \(A\). \(DCA\) and \(DOB\) are straight lines. It is given that angle \(BAR = 3 \times \text{angle } OCB\), prove that

angle \(ODC = 2 \times \text{angle } OBC\),

triangle \(COD\) is an isosceles triangle, and

angle \(DCO = \text{angle } CAT + 90^\circ\).

Properties of Special Triangles

<p>In the diagram, \(ABCD\) is a cyclic quadrilateral. \(AB\) is a diameter and is parallel to \(CD\). \(TDS\) and \(TBR\) are tangents to the circle at \(D\) and \(B\), respectively. Given that&nbsp;\(TDS\) intersects \(OA\) produced at \(S\), prove that</p>
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<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">angle \(ADS =\) angle \(TBC\) ,</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">angle \(CBD = 90^\circ - 2 \times\) angle \(SDA\) , and</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">triangles \(SAD\) and \(BCD\) are similar.</div></li></ol>

In the diagram, \(ABCD\) is a cyclic quadrilateral. \(AB\) is a diameter and is parallel to \(CD\). \(TDS\) and \(TBR\) are tangents to the circle at \(D\) and \(B\), respectively. Given that \(TDS\) intersects \(OA\) produced at \(S\), prove that

angle \(CBD = 90^\circ - 2 \times\) angle \(SDA\) , and

triangles \(SAD\) and \(BCD\) are similar.

Angles in Cyclic Quadrilaterals

<p>In the diagram, \(SA\) is the tangent to the circle with centre \(O\) at \(A\) and \(AB\) is a diameter. \(C\) is a point on the circle and the tangent to the circle at \(C\) intersects \(SA\) at \(T\). \(BCS\) is a straight line. Prove that</p>
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<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">angle \(CTS = 2 \times\) angle \(CAT\) ,</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">\(ST = CT\) , and</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">\(SC \times OC = AC \times AT\) .</div></li></ol>

In the diagram, \(SA\) is the tangent to the circle with centre \(O\) at \(A\) and \(AB\) is a diameter. \(C\) is a point on the circle and the tangent to the circle at \(C\) intersects \(SA\) at \(T\). \(BCS\) is a straight line. Prove that

<p>In the diagram, \(O\) is the centre of the circle, \(AC\) is a diameter and \(PAQ\) is the tangent to the circle at \(A\). Prove that</p>
<div class="graph-block" data-graph-config="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" data-width="223" data-height="173" data-proportional style="width:223px;height:173px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:223px;--graph-aspect:223 / 173;"></div>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">triangles \(CAP\) and \(CBA\) are similar, and</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">\(AC^2 = CP \times CB\).</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">State a triangle that is similar to triangle \(CAQ\). Hence, or otherwise, prove that&nbsp;\(CP \times CB = CD \times CQ\).</div></li></ol>

In the diagram, \(O\) is the centre of the circle, \(AC\) is a diameter and \(PAQ\) is the tangent to the circle at \(A\). Prove that

triangles \(CAP\) and \(CBA\) are similar, and

State a triangle that is similar to triangle \(CAQ\). Hence, or otherwise, prove that \(CP \times CB = CD \times CQ\).

<p>In the figure, \(O\) is the centre of the circle, \(AC\) is a diameter and \(PAQ\) is a tangent to the circle at \(A\).</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><ol class="i-list"><li><div class="list-item-paragraph">Show that&nbsp;\(\Delta CAP\) is similar to \(\Delta CBA\).</div></li><li><div class="list-item-paragraph">Prove that&nbsp;\(AC^2=CP \times CB\).</div></li><li><div class="list-item-paragraph">State a triangle that is similar to \(\Delta CAQ\).</div></li></ol><p>Hence, or otherwise, prove that&nbsp;\(CP \times CB = CD \times CQ\).</p></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure alignright" style="text-align: right;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1768112050469-d7q2dds3l0h.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMTIwNTA0NjktZDdxMmRkczNsMGgucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTYzNiwiZXhwIjoxNzkyNzI1OTM2fQ.IH5SqowZdqh2qIm76WBidNFVRrhefEDrBQGP93WWo2w" style="max-width: min(200px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

In the figure, \(O\) is the centre of the circle, \(AC\) is a diameter and \(PAQ\) is a tangent to the circle at \(A\).

<p>In the figure, \(TA\) and \(TB\) are tangents to the circle, centre \(O\), from \(T\).</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2-1" data-layout="2-1"><div class="question-column" style="flex: 2; align-self: flex-start;"><ol class="i-list"><li><div class="list-item-paragraph">Given that&nbsp;\(AOC\) is a diameter of the circle, state a reason why \(\angle AOB=2\angle ACB\).</div></li></ol><p>Hence, prove that</p><ol class="i-list" start="2"><li><div class="list-item-paragraph">\(CB\) is parallel to \(OT\),</div></li><li><div class="list-item-paragraph">A circle with diameter \(OT\) can be drawn to pass through \(A\) and \(B\),</div></li><li><div class="list-item-paragraph">\(\Delta OBC\) is similar to \(\Delta TAB\).</div></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure alignright" style="text-align: right;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1768110672714-c5voub3uhv7.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMTA2NzI3MTQtYzV2b3ViM3VodjcucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTYzNiwiZXhwIjoxNzkyNzI1OTM2fQ.YyUqHwKAnfrkfY4lfXTAfm98Kh4ET-mQmaDBGksUQwA" style="max-width: min(200px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

In the figure, \(TA\) and \(TB\) are tangents to the circle, centre \(O\), from \(T\).

<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><p style="text-align: left">The diagram shows a circle \(ABC\), with centre \(O\).<br>\(AP\) and \(BP\) are tangents to the circle.<br>\(\angle OAB = 25{}^\circ\)</p><ol class="abc-list"><li data-list-item-id="item-1780222062457-nj33mtt6n" data-list-item-path="item-1780222062457-nj33mtt6n"><div class="list-item-paragraph">Stating your reasons clearly, find</div><ol class="i-list"><li data-list-item-id="subitem-1780222066080-ekmhsk9a3" data-list-item-path="item-1780222062457-nj33mtt6n/subitem-1780222066080-ekmhsk9a3"><div class="list-item-paragraph">\(\angle AOB\).</div></li><li data-list-item-id="subitem-1780222066212-guqtj6m6s" data-list-item-path="item-1780222062457-nj33mtt6n/subitem-1780222066212-guqtj6m6s"><div class="list-item-paragraph">\(\angle PAB\).</div></li><li data-list-item-id="subitem-1780222066397-m8ji1u9yw" data-list-item-path="item-1780222062457-nj33mtt6n/subitem-1780222066397-m8ji1u9yw"><div class="list-item-paragraph">\(\angle APB\).</div></li></ol></li><li data-list-item-id="item-1780222063624-p3k2ioy91" data-list-item-path="item-1780222063624-p3k2ioy91"><div class="list-item-paragraph">For this part, use a separate configuration with the same tangency relationships, without the \(25^\circ\) condition from part (a). Given radius \(5\) cm and \(OP=16\) cm, find \(AP\).</div></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure alignright" style="text-align: right;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1767602723698-5zgahl0huoa.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njc2MDI3MjM2OTgtNXpnYWhsMGh1b2EucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTYzNiwiZXhwIjoxNzkyNzI1OTM2fQ.g8hNXssieJfFavR7PX_CmdmEbN05TkEIv7p4B0EVL6g" style="max-width: min(250px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

Applications of Pythagoras’ Theorem

Solving for Variables in Intersecting Circle Lines and Tangents

Properties of Circle

<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><p>In the diagram below, \(O\) is the centre of the circle and points \(A\), \(B\), \(C\) and \(D\) lie on the circumference of the circle.<br>The tangent at \(A\) meets \(CD\) produced at \(E\).<br>The line \(AC\) passes through \(O\) and lines \(AC\) intersects \(BD\) at point \(F\).<br>It is also given that&nbsp;\(\angle ACD=46{}^\circ\) and \(\angle CAB=25{}^\circ\).<br>Stating your reasons clearly, find</p><ol class="abc-list"><li><div class="list-item-paragraph">\(\angle ADO\).</div></li><li><div class="list-item-paragraph">\(\angle OFD\).</div></li><li><div class="list-item-paragraph">\(\angle AEC\).</div></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1767602520657-fzpds21ague.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njc2MDI1MjA2NTctZnpwZHMyMWFndWUucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTYzNiwiZXhwIjoxNzkyNzI1OTM2fQ.NqGzcPTVEHXW5wGaFBSFF5hYnxCag8JUceLYwQO-Hds" style="max-width: min(300px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

Angles in the Same Segment

The Geometry of Inscribed Triangles and Cyclic Polygons

<p>In the figure, \(AE\) is a tangent to the circle whose centre is \(O\) and \(ABC\) is a straight line.<br>\(\angle BAE = 38{}^\circ\), \(\angle BOE = 72{}^\circ\) and \(\angle DCE = 40{}^\circ\).<br>Find, stating your reasons clearly</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><ol class="abc-list"><li><div class="list-item-paragraph">\(\angle ECB\),</div></li><li><div class="list-item-paragraph">\(\angle OEC\),</div></li><li><div class="list-item-paragraph">\(\angle CED\),</div></li><li><div class="list-item-paragraph">\(\angle OBA\).</div></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure alignright" style="text-align: right;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1767602304815-a9qazn86x5.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njc2MDIzMDQ4MTUtYTlxYXpuODZ4NS5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyNzI1NjM2LCJleHAiOjE3OTI3MjU5MzZ9.j8pHGUt46tN60HrTnFbUKxkDmB2nQgbiJ6-kbxZSKS8" style="max-width: min(250px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

In the figure, \(AE\) is a tangent to the circle whose centre is \(O\) and \(ABC\) is a straight line.
\(\angle BAE = 38{}^\circ\), \(\angle BOE = 72{}^\circ\) and \(\angle DCE = 40{}^\circ\).
Find, stating your reasons clearly

Determining Angular Values from Diameters, Chords, and Arcs

<p>In the diagram, \(A\), \(B\), \(C\) and \(D\) are points on the circumference of the circle centre \(O\).<br>\(ABE\) and \(DCE\) are straight lines and \(ST\) is a tangent to the circle at \(A\).<br>\(ST\) is parallel to \(DE\), \(\angle CEB=36{}^\circ\) and \(\angle DAB=83{}^\circ\).</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><ol class="abc-list"><li><div class="list-item-paragraph">Calculate</div><ol class="i-list"><li><div class="list-item-paragraph">angle \(OAB\),</div></li><li><div class="list-item-paragraph">angle \(ABC\),</div></li><li><div class="list-item-paragraph">angle \(AOD\).</div></li></ol></li><li><div class="list-item-paragraph">Show that&nbsp;\(ACD\) is an isosceles triangle.</div></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1767601914948-dxhsqkds96b.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njc2MDE5MTQ5NDgtZHhoc3FrZHM5NmIucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTYzNiwiZXhwIjoxNzkyNzI1OTM2fQ.Ahcekz5rIzmCeFrZVkKrc7zoQnxKmlq03vF10tZCWf4" style="max-width: min(400px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

In the diagram, \(A\), \(B\), \(C\) and \(D\) are points on the circumference of the circle centre \(O\).
\(ABE\) and \(DCE\) are straight lines and \(ST\) is a tangent to the circle at \(A\).
\(ST\) is parallel to \(DE\), \(\angle CEB=36{}^\circ\) and \(\angle DAB=83{}^\circ\).

Investigating Geometric Proofs through Circle Angle Invariants

<p>In the diagram, \(O\) is the centre of the circle through \(A\), \(B\), \(C\) and \(D\) and \(TB\) is the tangent to the circle at \(B\).<br>The diameter \(AC\) and the chord \(BD\) intersects at \(X\).<br>Given&nbsp;\(\angle DCX = 32{}^\circ\) and \(\angle OBC = 40{}^\circ\), calculate, stating your reasons clearly,</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><ol class="abc-list"><li><div class="list-item-paragraph">\(\angle OAB\),</div></li><li><div class="list-item-paragraph">\(\angle CXD\),</div></li><li><div class="list-item-paragraph">\(\angle ABT\),</div></li><li><div class="list-item-paragraph">\(\angle BAD\).</div></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure alignright" style="text-align: right;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1767601627209-u1wmorjuuwo.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njc2MDE2MjcyMDktdTF3bW9yanV1d28ucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTYzNiwiZXhwIjoxNzkyNzI1OTM2fQ.RYPseF7XRiE86adm3wStB0yorJMA-1DKTl-_TDHb5d4" style="max-width: min(250px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

In the diagram, \(O\) is the centre of the circle through \(A\), \(B\), \(C\) and \(D\) and \(TB\) is the tangent to the circle at \(B\).
The diameter \(AC\) and the chord \(BD\) intersects at \(X\).
Given \(\angle DCX = 32{}^\circ\) and \(\angle OBC = 40{}^\circ\), calculate, stating your reasons clearly,

Euclidean Geometry: Interior Angles and Circular Arc Relationships

<p>\(TA\) and \(TB\) are tangents to a circle centre \(O\) and \(AT\) produced meets \(OB\) at \(Y\).<br>\(YB=12\) cm, \(TB=5\) cm and \(OA=r\) cm.</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><ol class="abc-list"><li><div class="list-item-paragraph">Write down the length of \(TA\).</div></li><li><div class="list-item-paragraph">Calculate the length of \(TY\).</div></li><li><div class="list-item-paragraph">Without solving <strong>(d)</strong>, find the numerical value of \(\frac{OA}{AY}\).</div></li><li><div class="list-item-paragraph">Calculate the value of \(r\).</div></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1780571603336-2ndg69la94u.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3ODA1NzE2MDMzMzYtMm5kZzY5bGE5NHUucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTYzNiwiZXhwIjoxNzkyNzI1OTM2fQ.2PkeUN1yCmmoZKociowXCqtnHtG2D0AmEvvNXQYa91c" style="max-width: min(350px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

\(TA\) and \(TB\) are tangents to a circle centre \(O\) and \(AT\) produced meets \(OB\) at \(Y\).
\(YB=12\) cm, \(TB=5\) cm and \(OA=r\) cm.

Calculation of Unknown Lengths in Circular Segments

<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><p>In the diagram, \(PA\) and \(PB\) are tangents to the circle with centre, \(O\).<br>\(PCO\) is a straight line where&nbsp;\(C\) is a point on the circle.<br>Find the</p><ol class="abc-list"><li><div class="list-item-paragraph">value of \(m\), where&nbsp;\(PC=m\) cm.</div></li><li><div class="list-item-paragraph">angle \(BPA\).</div></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure alignright" style="text-align: right;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1767597257539-1jh37mgbrz8.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njc1OTcyNTc1MzktMWpoMzdtZ2JyejgucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTYzNiwiZXhwIjoxNzkyNzI1OTM2fQ.KdWq0XiClEIPsoj9xWOMxiNHKFzpwwln5avWWyd2I-I" style="max-width: min(280px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

Basics of Trigonometric Ratios

Geometry Paper: Circle Theorems and Applications

Trigonometry

<div class="content-block content-block--columns"><div class="question-columns question-columns--2-1" data-layout="2-1"><div class="question-column" style="flex: 2; align-self: flex-start;"><p>The diagram shows a circle, centre \(O\) and radius \(3\) cm.<br>If&nbsp;\(CA\) and \(CB\) are tangents to the circle at \(A\) and \(B\) respectively and \(AC=8\) cm, find</p><ol class="abc-list"><li><div class="list-item-paragraph">the length of \(BC\),</div></li><li><div class="list-item-paragraph">the length of \(OC\),</div></li><li><div class="list-item-paragraph">the area of the quadrilateral \(AOBC\).</div></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1767595729883-e98jjobh0d7.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njc1OTU3Mjk4ODMtZTk4ampvYmgwZDcucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTYzNiwiZXhwIjoxNzkyNzI1OTM2fQ.9MGQNcZnWERpxEx3ES830HBsRh4zEIogY_O_7zB65Zc" style="max-width: min(300px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

Area of Triangle

Applications of Circle Symmetric Property 4

Perimeter and Area of Plane Figures

<div class="content-block content-block--columns"><div class="question-columns question-columns--2-1" data-layout="2-1"><div class="question-column" style="flex: 2; align-self: flex-start;"><p style="text-align: left">\(O\) and \(P\) are the centres of \(2\) circles, with radii \(3\) cm and \(3.5\) cm respectively.<br>The two circles touch externally at \(Q\), and \(OM\) is a tangent to the larger circle at \(N\).<br>Calculate the length of \(ON\).</p></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1767595544827-h0q6zuwxbf5.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njc1OTU1NDQ4MjctaDBxNnp1d3hiZjUucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTYzNiwiZXhwIjoxNzkyNzI1OTM2fQ.BgthfSeTwBP-iYaQJJYsgVeJSQne25B2XgXfYY1y7OI" style="max-width: min(300px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

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